Question

Two identical conducting rods are first connects independently to two vessels, one containing water at 100 ${}^0$C and the other containing ice at 0 C . In the second case, the rods are joined end to end and connected to the same vessels. Let $q_1$ and $q_{{}_2}$ g/s be the rate of the melting of ice in the two cases respectively . the ratio $\frac{q_1}{q_2}$ is:

• A. $\frac{1}{2}$
• B. $\frac{2}{1}$
• C. $\frac{4}{1}$
• D. $\frac{1}{4}$

Answer 1

The correct option is D $\frac{1}{4}$

In the first case, the two roads are joined in center.

[$k$ : conductance of rod]

[$\Delta T$ : Temperature differences] [Ref. image 1]

$q_1$ is the rate of reat transfer through rod.

Now, equivalent conduction once in series,

$K_1=\frac{k}{2}$ $[using\frac{1}{k_1}=\frac{1}{k}+\frac{1}{k}]$

So as we get reat trnasfer rate,

$q_1=\Delta K_1\Delta T=\frac{K}{2}\Delta T$ ....(1)

In second case, the rods are joined in parallel. [Ref. image 2]

Equivalent conductance in parallel,

$K_2=2K$

Thus, we get:

$q_2=K_2\Delta T=2K\Delta T$ ...(2)

Taking ratio $\left(1\right)÷\left(2\right)$,

$\frac{q_1}{q_2}=\frac{K_1\Delta T}{2.2K_2\Delta T}\Rightarrow q_1:q_2=1:4$

Note: One can think the rod as resistances 'R' applied with LOOV; once is series and once in parallel, and take ratio of currents, from each case.

Expression $Q=K\Delta T$ is siliar to ohm's law, $I=GV[G=\frac{1}{K}]$